How do you solve #3x^2 + 10x - 2 = 0 # by completing the square?

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Answer 1 · 2017-10-07T06:56:55

#x=sqrt31/9-5/3#
#x=-sqrt31/9-5/3#

Given -

#3x^2+10x-2=0#

Take the constant term to right-hand side

#3x^2+10x=2#

Divide both sides by 3

#(3x^2)/3+(10x)/3=2/3#

#x^2+10/3x=2/3#

Divide the coefficient of #x# by2; square it and add it to both sides

#x^2+10/3x+100/36=2/3+100/36=(24+100)/36=124/36=31/9#
#(x+10/6)^2=31/9#

#x+5/3=+-sqrt(31/9)= +-sqrt31/3#

#x=sqrt31/9-5/3#
#x=-sqrt31/9-5/3#